A 2.00-kg rock has a horizontal velocity of magnitude 12.0 ms wh...

Question

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.

(b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?

Accepted Answer

Hey everyone in this problem, you're strolling around one evening when a 50 g marble is launched horizontally at 15 m per second from a 12 m high point in a tall building That's gonna be considered .8. Wait is the only force acting on the stone during the flight and were asked to calculate the magnitude and direction of the rate of change of the stones angular momentum about your feet, which is point B the moment it is launched. All right. So we're giving this diagram here. We have us standing at point B 20 m away is the base of the house. The house is 12 m tall and at the top, we have .8 From point a this marble is being launched horizontally to the left at 15 m/s. Our answer choices are a 11. kg meter squared per second squared. Out of the page B 5.88 kg meters squared per second squared out of the page C 9. kg meter squared per second squared out of the page D 11.4 kg meter squared per second squared into the page E 5.88 kg meters squared per second squared into the page or F 9. kg meter squared per second squared into the page. Now, what we're looking for is the change the rate of change of the angular momentum. Okay. So that's gonna be D L by D T, the rate of change of the angular momentum with respect to time. Now recall the D L by D T. Well, this is just equal to our torque tau. So if we want to calculate D L D T, we can just calculate our torque tau. How can we calculate our torque? Well, it's equal to our best sign of data. So let's draw another little diagram. So we have our point here where we're standing point B To the right of us, 20 m with the base of the house, We go up 12 m from that and we get point A. Now when we're talking about our, we're talking about the distance between the point of reference, which is our feet at B and the point of motion, which is the marble at point A. And so our is going to be this distance of the vector like connects the two, our forces acting straight downwards because we're told that weight is the only force acting on the stone. We know that that act straight downwards and our angle theta is gonna be the angle between our. So it's gonna be the angle in the top right corner of the triangle. Okay at point A. All right. So let's start with our force f. Okay. Our force f we're told weight is the only force acting and recall that the weight is equal to the mass times the gravitational acceleration. The mass of our marble is 50 g. And we want to convert this into kg are standard unit. So we're gonna multiply by one kg per 1000 g. Okay. We know there's one kg 1000 g. The unit of Graham will cancel, We're effectively dividing by 1000. And then we're gonna multiply by the gravitational acceleration 9.8 m/s squared. And if we work this out 50 divided by 1000 times 9. m per second squared, we get 0.49 kilogram meters per second squared. OK? And a kilogram meter per second squared, we know that that is equivalent to a Newton. So we have 0.45 Newton, which is a unit we would want for four. So the units check out here. Now we have our force F. We also need to find our and signed data. Now we're gonna do a little trick here. R and theta are both part of the same triangle. So instead of calculating are separately and then sine theta separately, we can actually calculate them together if we consider in our triangle, this angle. And let's consider sine of that angle because that's what we want in our D L D T equals torque equation. Okay, let's consider sign of fate. Now sign of data is going to be equal to the opposite side which is m divided by the high produce, which is our. Okay. Well, we can multiply both sides by our and this is going to give us R sine theta is equal to 20 m. So now instead of finding our and then finding signed data, multiplying them, we know that our time signed data is 20 m and that's going to save us a couple of steps in our calculation. Alright, so getting back to what we're looking for the change rate of change of the angular momentum D L D T is equal to R F sine theta, we can write this as F times are signed, these terms are multiplied together so we can rearrange them. However, we'd like we found that our force was equal to 0.49 Newtons And the term r sine Theta was equal to 20 m. And so we get DLDT is .49 Newtons times m, which gives us 9.8 Newton meters. Right now, we know that a Newton is a kilogram meter per second squared. And so 9.8 Newton meters is 9.8 kg meters squared per second squared, the unit that we have in those answer choices. Alright, so we found the magnitude of the rate of change, okay. 9.8 kg meter squared per second squared. So we're looking at option choice either C or F. Now, we need to figure out the direction and we're gonna use our right hand rule. We go back up to our diagram to use our right hand rule. We're gonna take our right hand, we're gonna point our fingers in the direction of our okay and then curl them in the direction of and your thumb is gonna point downwards into the page. When you do that, be careful here, we're curling in the direction of F not in the direction of the velocity V when we're talking about this torque. Okay. So pointing them in the direction of our curling your fingers in the direction of F our thumb points into the page. And so the direction is into the page and we have answer choice F and the magnitude of the rate of change of the angular momentum is 9.8 kg meter squared per second squared in the direction is into the page. Thanks everyone for watching. I hope this video helped see you in the next one.

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?

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A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.
(b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?